Problem: $\dfrac{d}{dx}[-4\sin(x)+9x]=$
Explanation: The expression to differentiate includes $\sin(x)$. Remember that the derivative of $\sin(x)$ is $\cos(x)$. Put another way, $\dfrac{d}{dx}[\sin(x)]=\cos(x)$. $\begin{aligned} &\phantom{=}\dfrac{d}{dx}[-4\sin(x)+9x] \\\\ &=-4\dfrac{d}{dx}[\sin(x)]+9\dfrac{d}{dx}(x) \\\\ &=-4\cdot \cos(x)+9\cdot 1 \\\\ &=-4\cos(x)+9 \end{aligned}$ In conclusion, $\dfrac{d}{dx}[-4\sin(x)+9x]=-4\cos(x)+9$